Question

Spin only magnetic moment of $Ni$ in $\left[Ni\right(dmg{)}_2]$ is same as that founds in :

• A. $Ni\text{in}\left[NiC{l}_2\right(PP{h}_3{)}_3]$
• B. $Mn\text{in}[Mn{O}_4{]}^{2-}$
• C. $Co\text{in}[CoB{r}_4{]}^{2-}$
• D. $Pt\text{in}\left[Pt\right(H_{2}O{)}_{2}B{r}_2]$

Answer 1

The correct option is D $Pt\text{in}\left[Pt\right(H_{2}O{)}_{2}B{r}_2]$

In $\left[Ni\right(dmg{)}_2]$ complex, the oxidation state of $Ni$ is +2. Nickel can form octahedral, tetrahedral and square planar comple. Here, $N{i}^{2+}$ forms square planar complex due to the pairing of electrons by strong filed (dmg) ligand. Thus, it has no unpaired electron and it is diamagnetic in nature.

in $Pt\text{in}\left[Pt\right(H_{2}O{)}_{2}B{r}_2]$, the platinum is +2 oxidation state. $Pt\left(II\right)$ always prefer inner orbital complex. Thus, it for square planar complex and it diamagnetic.
$Pt\left(II\right)$ have $5d^8$ configuration and complex is square planar and therefore, diamagnetic, So $\mu =0.$